Exercises — Kepler's connection — orbits are ellipses (motivation)

WHAT: average the two extremes. WHY: the two extreme points sit at opposite ends of the long axis, so half their sum is exactly the half-length $a$. $a=\frac{1}{2}(3+7)=\frac{1}{2}(10)=\boxed{5\text{ AU}}$

WHY this formula: recall $c$ is the focus's offset from the centre. The difference $r_{\max }-r_{\min }=a(1+e)-a(1-e)=2ae=2c$ measures twice that offset; dividing by the sum $r_{\max }+r_{\min }=2a$ turns it into a pure ratio $c/a=e$. $e=\frac{7-3}{7+3}=\frac{4}{10}=\boxed{0.4}$

WHY $a^{3/2}$: starting from $T^2=a^3$, take the square root of both sides: $T=\sqrt{a^3}=a^{3/2}$. $T=4^{3/2}=\sqrt{4^3}=\sqrt{64}=\boxed{8\text{ yr}}$

WHAT: plug into the min/max formulas. WHY: we are given the orbit's size $a$ and its off-centre-ness $e$, and want the two turning points. $r_{\min }=a(1-e)=10(1-0.9)=10(0.1)=\boxed{1\text{ AU}}$ $r_{\max }=a(1+e)=10(1+0.9)=10(1.9)=\boxed{19\text{ AU}}$ Check: $\frac{1}{2}(1+19)=10=a$ ✅. The orbit dives close then swings far out — that big $e$ means a long, thin loop.

WHY $b^2=a^2(1-e^2)$: use the coordinate setup from the opening figure — centre at the origin, foci on the $x$-axis at distance $c$ from centre. The top point of the ellipse is $(0,b)$. By left–right symmetry it is the same distance from each focus, and that common distance equals $a$ (the sum of the two equal distances is $2a$, so each is $a$). The centre, that top point, and one focus form a right triangle with legs $c$ (horizontal) and $b$ (vertical) and hypotenuse $a$: so $c^2+b^2=a^2$. Since $c=ae$, we get $b^2=a^2-a^2{e}^2=a^2(1-e^2)$. $b^2=10^2(1-0.9^2)=100(1-0.81)=100(0.19)=19$ $b=\sqrt{19}\approx \boxed{4.36\text{ AU}}$ Notice $b\ll a$: the orbit really is squashed.

WHY use the ratio form: the proportionality $T^2\propto a^3$ has an unknown constant, but taking a ratio of two planets around the same star cancels it. $\frac{T_Y^2}{T_X^2}=\frac{a_Y^3}{a_X^3}=\frac{9^3}{1^3}=729\Rightarrow \frac{T_Y}{T_X}=\sqrt{729}=\boxed{27}$

WHAT: compare $r_{\max }$ to $r_{\min }$ as a percentage. WHY a ratio: "by what percentage larger" means $(r_{\max }-r_{\min })/r_{\min }\times 100\%$, and here the $a$ cancels nicely. $\frac{r_{\max }}{r_{\min }}=\frac{a(1+e)}{a(1-e)}=\frac{1+e}{1-e}=\frac{1.0167}{0.9833}\approx 1.0340$ So aphelion is about $\boxed{3.40\%}$ larger than perihelion. Even though $e$ is only $0.0167$, the distance gap is roughly $2e\approx 3.3\%$ — small, which is why Earth's orbit looks like a circle.

WHY $rv$ is constant at the ends: the swept-area rate is $\frac{dA}{dt}=\frac{1}{2}{r}^2\dot{\theta }$, constant. At perihelion and aphelion the velocity is purely sideways (perpendicular to $r$), so $v=r\dot{\theta }$ there, giving $\frac{dA}{dt}=\frac{1}{2}rv$. Constant area rate ⇒ $r_p{v}_p=r_a{v}_a$. This is Angular momentum conservation in disguise. $v_a=\frac{r_p{v}_p}{r_a}=\frac{2\times 40}{8}=\frac{80}{8}=\boxed{10\text{ km/s}}$ The planet is $4\times$ farther, so it moves $4\times$ slower — exactly the "faster near the Sun" rule.

WHAT: first recover $a$ from $r_{\min }$ and $e$, then use Kepler 3. WHY: we can't use $a=\frac{1}{2}(r_{\min }+r_{\max })$ because $r_{\max }$ isn't given — but $r_{\min }=a(1-e)$ contains $a$ directly. $a=\frac{r_{\min }}{1-e}=\frac{0.5}{1-0.98}=\frac{0.5}{0.02}=25\text{ AU}$ $T=a^{3/2}=25^{3/2}=(\sqrt{25}{)}^3=5^3=\boxed{125\text{ yr}}$

Strategy — chain three ideas: (1) Kepler 3 gives $a$ from $T$; (2) $r_{\min }=a(1-e)$; (3) $r_{\max }=a(1+e)$. Step 1 — invert $T=a^{3/2}$ to get $a=T^{2/3}$: $a=8^{2/3}=(\sqrt[3]{8}{)}^2=2^2=4\text{ AU}$ Step 2 — perihelion: $r_{\min }=4(1-0.25)=4(0.75)=\boxed{3\text{ AU}}$ Step 3 — aphelion: $r_{\max }=4(1+0.25)=4(1.25)=\boxed{5\text{ AU}}$

WHY start from $b^2=a^2(1-e^2)$: we know both axes but not $e$, and this is the only formula linking $a,b,e$. Solve it for $e$. $b^2=a^2(1-e^2)\Rightarrow 1-e^2=\frac{b^2}{a^2}=\frac{16}{25}\Rightarrow e^2=1-\frac{16}{25}=\frac{9}{25}$ $e=\frac{3}{5}=\boxed{0.6}$ Then perihelion: $r_{\min }=a(1-e)=5(1-0.6)=5(0.4)=\boxed{2\text{ AU}}$ Sanity: $c=ae=3$, and $c^2+b^2=9+16=25=a^2$ ✅ — the right triangle closes.

Eccentricities (from extremes): $e_A=\frac{3-1}{3+1}=\frac{2}{4}=0.5,e_B=\frac{6-2}{6+2}=\frac{4}{8}=0.5$ Same shape! Both have $\boxed{e=0.5}$ — B is just a scaled-up copy of A. Periods (from $a$, via Kepler 3): $a_A=\frac{1}{2}(1+3)=2,a_B=\frac{1}{2}(2+6)=4$ $\frac{T_B^2}{T_A^2}=\frac{a_B^3}{a_A^3}=\frac{4^3}{2^3}=\frac{64}{8}=8\Rightarrow \frac{T_B}{T_A}=\sqrt{8}=2\sqrt{2}\approx \boxed{2.83}$ Same eccentricity, but B's year is about $2.83\times$ longer because it's a bigger orbit.

Average: $\frac{r_{\min }+r_{\max }}{2}=\frac{a(1-e)+a(1+e)}{2}=\frac{a-ae+a+ae}{2}=\frac{2a}{2}=a✓$ The $\pm ae$ terms cancel — that's why the average kills the offset and returns pure $a$. Product: $r_{\min }\cdot r_{\max }=a(1-e)\cdot a(1+e)=a^2(1-e)(1+e)=a^2(1-e^2)=b^2✓$ WHAT IT MEANS: the geometric mean of perihelion and aphelion is the semi-minor axis $b$, while the arithmetic mean is the semi-major axis $a$. Since arithmetic mean $\ge$ geometric mean, this instantly re-proves $a\ge b$.

Step 1 — use equal areas ($r_p{v}_p=r_a{v}_a$) to get $r_{\max }$: $r_{\max }=\frac{r_{\min }{v}_p}{v_a}=\frac{1\times 60}{20}=3\text{ AU}$ WHY: the speed ratio is $3:1$, so the distance ratio must be $1:3$ — angular momentum $rv$ is conserved at the two ends. Step 2 — geometry from extremes: $a=\frac{1}{2}(1+3)=\boxed{2\text{ AU}},e=\frac{3-1}{3+1}=\boxed{0.5}$

Limit $e\to 0$ (circle): $r_{\min }=a(1-e)\to a,r_{\max }=a(1+e)\to a.$ Nearest = farthest = $a$: the two foci merge at the centre and the ellipse becomes a circle of radius $a$. Also $b^2=a^2(1-e^2)\to a^2$, so $b\to a$ — no squashing. Limit $e\to 1$ (needle): $r_{\min }=a(1-e)\to 0,r_{\max }=a(1+e)\to 2a.$ The planet skims arbitrarily close to the Sun and swings out to nearly $2a$. Also $b^2=a^2(1-e^2)\to 0$, so $b\to 0$: the ellipse collapses to a line segment. And exactly at $e=1$ the shape is no longer a closed ellipse at all — it opens into a parabola (see Parabola — projectile motion connection); for $e>1$ it becomes an unbound hyperbola (Hyperbola — unbound orbits and escape) — the object escapes and never returns. The full family: $e=0$ circle → $0<e<1$ ellipse → $e=1$ parabola (escape, just barely) → $e>1$ hyperbola. One number $e$ names them all — the heart of Eccentricity of a conic and the Focus–directrix definition of conics.

• From $T=8$ yr, what is $a$? ::: $a=T^{2/3}=8^{2/3}=4$ AU.
• Product $r_{\min }{r}_{\max }$ equals what? ::: $b^2$ (semi-minor axis squared).
• Speed ratio perihelion:aphelion when $r$ ratio is $1:4$? ::: $4:1$ (from $rv$ constant).
• As $e\to 1$ with $a$ fixed, $r_{\min }\to$? ::: $0$.
• Which $e$ gives a parabola? ::: exactly $e=1$.